# Replacing an operator of an angular momentum for a constant.

1. Jun 5, 2013

### 71GA

While dealing with a circling particle in an spherical symetric potential our professor said that we can replace an operator of $z$ component of angular momentum $\hat{L}_z$ with the expectation value - he denoted it just $L_z$ - of the angular momentum if $L_z$ is constant. Why is that?

So we first had this equation:

\begin{align}
\underbrace{\psi (r,\varphi,\vartheta)}_{\rlap{\text{w. f. in spherical coordinates}}} &= \exp\left[\hat{L}_z \frac{i}{\hbar}\, \varphi\right] \underbrace{\psi (r,0,\vartheta)}_{\rlap{\text{w. f. in spherical. coordinates at $\varphi=0$}}}
\end{align}

and we got this one (notice that there is no operator over an $L_z$):

\begin{align}
\psi (r,\varphi,\vartheta) &= \exp\left[L_z \frac{i}{\hbar}\, \varphi\right] \psi (r,0,\vartheta)
\end{align}

Anyway here is the spherical coordinate system we ve been using all the time (the blue spherical aure is supposed to be a spherical potential...):

Last edited: Jun 5, 2013
2. Jun 5, 2013

### Jano L.

I think there should be imaginary unit inside the[ ] brackets. The first equation states that the function $\psi$ is analytic in $\varphi$. The second equation states more than that: it says that the function behaves as $e^{im\varphi}$, i.e. oscillates. Such functions are eigenfunctions of the angular momentum operator, if the number $m = l_z/\hbar$ is whole number.

3. Jun 5, 2013

### 71GA

You are soo right! I misstyped $i$ for $1$. I corrected that mistake. I don't think i understand your explaination though.